The Definition of B — Question 6

Question 6

Reversing a test velocity separates electric and magnetic contributions

A charge q=2.00 nanocoulomb moves first with $+1000\hat x$ m/s and then with $-1000\hat x$ m/s through unchanged electric and magnetic fields. The forces are $(4\hat x+6\hat y)$ nN and $(4\hat x-2\hat y)$ nN respectively.

1. Find E from the even-in-velocity part of the force.

2. Find the magnetic-field components accessible to the experiment and identify the undetermined component.

Question 6 - Solution

Reversing a test velocity separates electric and magnetic contributions

(a) Method and calculation

$$\vec F_++\vec F_-=2q\vec E\Rightarrow\vec E=\frac{(8\hat x+4\hat y)10^{-9}}{4\times10^{-9}}\,\mathrm{V/m}.$$$$\vec E=(2\hat x+\hat y)\,\mathrm{V/m}.$$

(b) Interpretation and further calculation

$$\tfrac12(\vec F_+-\vec F_-)=4\hat y\,\mathrm{nN}=qv(-B_z\hat y+B_y\hat z).$$$$B_z=-2.00\,\mathrm{mT},\quad B_y=0,\quad B_x\text{ undetermined}.$$

Physical check

Electric force is independent of the test velocity; magnetic force is odd under its reversal.